Question

The mean and variance of Binomial random variable X are respectively 4 and 1. Then probability that X takes the value greater than 1 is

• A. ${\left(\frac{3}{4}\right)}^{16}$
  
• B. ${\left(\frac{3}{4}\right)}^{14}.\frac{1}{4}$
  
• C. $1-\frac{47}{4^{16}}$
  
• D. $1-\frac{49}{4^{16}}$

Answer 1

The correct option is D $1-\frac{49}{4^{16}}$

Given np =4, npq =1 $\Rightarrow q=\frac{1}{4}\Rightarrow p=\frac{3}{4}\Rightarrow p=\frac{3}{4}\Rightarrow n=16$
$\therefore$ Binomial distribution is $p(x=r){=}^{16}{C}_r{p}^r{q}^{16-r}r=0,1,2,....16$
Required probability = p(x > 1)
= 1 - p(x < 1)
= 1 - [p(x = 0) + p(x = 1)]
$=1-[q^{16}+16p{q}^{15}]$
$=1-[q+16p]q^{15}$
$=1-[\frac{1}{4}+16.\frac{3}{4}]{\left(\frac{1}{4}\right)}^{15}=1-\frac{49}{4}.{\left(\frac{1}{4}\right)}^{15}=1-\frac{49}{4^{16}}$